Lecture 20

Convergence proofs

Let’s take a brief detour to remind you of something that you probably know, but whose importance may not have been pointed out to you. The triangle inequality says that x+yx+y|x| + |y| \geq |x+y| for any real numbers xx and yy. It’s easy to prove, by carefully analysing what can happen: which combinations of signs of xx, yy and x+yx+y are possible?

We can use this to get the following: zy+yx(zy)+(yx)=zx.|z-y| + |y-x| \geq |(z-y)+(y-x)| = |z-x|. This embodies the following slogan:

The distance from xx to zz if we go direct is less than if we go via yy.

Now we get back to the subject of convergence.

We say that a sequence (ai)iN=a0,a1,(a_i)_{i\in\mathbb{N}} = a_0, a_1, \ldots is convergent if it converges to some xx.

Here’s a very important fact (which is only true because of all that work we put in finding a good definition):

Proposition

A sequence a0,a1,a_0, a_1, \ldots cannot converge to two different real numbers xx and yy.

Before we prove this, let’s think about what a proof might look like. Here’s a picture of a numberline, to help show us what would happen if a sequence was convergent to xx and to yy:

The two bars at the bottom of that diagram were deliberately chosen not to overlap (I made each of them extend one third of the distance from xx to yy, leaving another third of the distance between them in the middle).

And since it converges to xx and yy, eventually all the terms of the sequence should be within the interval around xx, and also all of them within the interval around yy, which is a contradiction since the intervals don’t meet.

Let’s do the working carefully.

Proof

We’ll prove this by contradiction. So, suppose it can: suppose that there is a sequence a0,a1,a_0, a_1, \ldots, which converges to two different real numbers xx and yy. Without loss of generality, we may take x<yx<y.

Since the sequence a0,a1,a_0,a_1,\ldots converges to xx, there is some NN such that, for all n>Nn>N, we have anx<yx3\left|a_n - x\right|<\frac{y-x}{3}.

Since the sequence a0,a1,a_0,a_1,\ldots converges to yy, there is some MM such that, for all n>Mn>M, we have any<yx3\left|a_n - y\right|<\frac{y-x}{3}.

But then, using the triangle inequality, for any nn bigger than both NN and MM, we have yx=yxyan+xan<yx3+yx3=23(yx)y-x=\left|y-x\right| {}\leq\left|y-a_n\right|+\left|x-a_n\right| {}<\frac{y-x}{3}+\frac{y-x}{3} {}=\frac{2}{3}(y-x) which is a contradiction as yxy-x is positive.

As a result, if a sequence is convergent, there is a unique real number to which it converges; we call that the limit of the sequence.

Let’s now try proving that some sequence or other does converge, as we’re not well practiced at that yet:

Proposition

The sequence a1=0,a2=1/2,a3=2/3,a4=3/4,a5=4/5,a_1=0,\quad a_2=1/2, \quad a_3=2/3, \quad a_4=3/4, \quad a_5=4/5,\quad\ldots where an=n1na_n = \frac{n-1}{n}, converges to 11.

Proof

[Proof (rough version)]

The definition of convergence is complicated, so it may be helpful to start by reminding us what we’re aiming for. So we’ll start by working from the wrong end.

We need to show that, for every ϵ>0\epsilon>0, there is some NN, such that for all n>Nn>N we have n1n1<ϵ.\left| \frac{n-1}{n} - 1 \right| < \epsilon.

A natural thing to do is to simplify that: n1n1=(n1)nn=1n=1n.\left|\frac{n-1}{n} - 1 \right| {= \left| \frac{(n-1) - n}{n}\right|} {= \left|\frac{-1}{n}\right|} {= \frac{1}{n}}.

So, as we can see, what we’re aiming for is that, for every ϵ>0\epsilon>0 there is some NN, such that for all n>Nn>N we have 1n<ϵ.\frac{1}{n} < \epsilon. But that’s the same as having n>1ϵ.n > \frac{1}{\epsilon}.

So if we take NN to be 1ϵ\left\lceil\frac{1}{\epsilon}\right\rceil, the smallest integer greater than 1/ϵ1/\epsilon, that works.

That proof is sort-of-okay, but it’s backwards. It was helpful to write it, but hard to check that it’s logically valid. I’ll now rewrite it forwards.

Proof

[Proof (neat version)] We must show that, for every ϵ>0\epsilon>0, there is some NN such that for all n>Nn>N we have n1n1<ϵ.\left| \frac{n-1}{n} - 1 \right| < \epsilon. Let such an ϵ\epsilon be given.

Define NN to be 1ϵ\left\lceil\frac{1}{\epsilon}\right\rceil, which is the smallest integer greater than 1/ϵ1/\epsilon.

Then, if n>Nn>N, we have n1n1=(n1)nn=1n=1n<1N<11/ϵ=ϵ,\left| \frac{n-1}{n} - 1 \right| {}= \left| \frac{(n-1) - n}{n}\right| {}= \left|\frac{-1}{n}\right| {}= \frac{1}{n} {}< \frac{1}{N} {}< {}\frac{1}{1/\epsilon} {}= \epsilon, exactly as required.

That second version is obviously correct, and all the reasoning goes in the right direction. But analysis proofs often have the property that the most persuasive proof will seem a bit mysterious. It’s best to do the rough work and then rewrite it neatly.

I understand you will have met the subject of convergence in MAS110.

That course is about streetfighting, and there you’re encouraged to use any techniques you have to hand without worrying too much about what it means.

This is a course about developing fundamental in mathematics and their proofs: if I set problems about convergence in MAS114, I need you to give a rigorous proof, with everything traced back to the definition of convergence (unless you’re told otherwise), rather than using the slightly vaguer methods and extra theorems you saw there!